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\begin{document}

\title{高等代数一}
\subtitle{8-习题与问答-线性方程组 }
%\institute{上海立信会计金融学院}
%\author{王立庆}
\author{{\ppr LQW}}
%\renewcommand{\today}{{\ppr \number\year \,年 \number\month \,月 \number\day \,日} }
\date{{\ppr 2022年10月18日} }

\maketitle

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%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{内容提要 }

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\begin{enumerate}

\item  克拉默公式、或线性方程组的解集的结构
\item  矩阵的初等变换
\item  矩阵的秩
\item  线性方程组有解的判别
\item  带参数的线性方程组

\end{enumerate}


\end{frame}

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%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{讲解本次作业的同学 }

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{\small 
\begin{table}[ht]
\centering
\begin{tabular}{cccccc}
4-习题&8-习题&12-习题&16-习题&20-习题&24-习题 \\ \hline 
{01}&\underline{02}&03&04&05&06 \\   
{07}&\underline{08}&09&10&11&12 \\  
{13}&\underline{14}&15&16&17&18 \\ 
{19}&\underline{20}&21&22&23&24 \\  
{25}&\underline{26}&27&28&29&30 \\  
{31}&\underline{32}&33&34&35&36 \\  
{37}&\underline{38}&39&40&41&42 \\  
{43}&\underline{44}&45&46&47&48 \\ 
\underline{49}&\underline{50}&51&52&53&54 \\  
\end{tabular}
\end{table}
}

\end{frame}

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%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{8.1. 克拉默公式、或线性方程组的解集的结构 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
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\begin{itemize}

\item  习题1：设 $a_1,a_2,a_3, a_4$ 是四个不同的实数，设 $b_1,b_2,b_3,b_4$ 是任意给定的实数。证明：存在唯一的实系数多项式 $f(x)=c_0+c_1x+c_2x^2+c_3x^3$, 使得 $f(a_i)=b_i$ 对 $i=1,2,3,4$ 都成立。

\item  解答思路：化为求解一个线性方程组。然后考察系数矩阵的秩。

\end{itemize}

\end{frame}

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%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{8.2. 矩阵的初等变换 }

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\begin{itemize}

\item  习题2：找到若干次第二类和第三类行初等变换，实现下述矩阵的变换，
{\footnotesize 
\begin{eqnarray*}
\begin{bmatrix} 
a_1&a_2&a_3&a_4 \\  
b_1&b_2&b_3&b_4 \\  
c_1&c_2&c_3&c_4 \\ 
\end{bmatrix}
\longrightarrow 
\begin{bmatrix} 
b_1&b_2&b_3&b_4 \\  
a_1&a_2&a_3&a_4 \\  
c_1&c_2&c_3&c_4 \\ 
\end{bmatrix}. 
\end{eqnarray*}
}

\item  解答思路：各种尝试。例如将第一行乘以 $-1$ 加到第二行。

\end{itemize}

\end{frame}

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%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{8.3. 矩阵的初等变换 }

\vspace{-0.4cm}\noindent\makebox[\linewidth]{\rule{\paperwidth}{0.4pt}}
%每页详细内容

\begin{itemize}

\item  习题3：设 $ad-bc\neq 0$. 证明：存在一系列的行初等变换，使得
{\footnotesize 
\begin{eqnarray*}
\begin{bmatrix} 
a&b \\ 
c&d \\ 
\end{bmatrix}
\longrightarrow 
\begin{bmatrix} 
1&0 \\ 
0&1 \\ 
\end{bmatrix}. 
\end{eqnarray*}
}

\item  解答思路：首先可知 $a,c$ 不全为零。然后分情况讨论：
\begin{enumerate}
\item  当 $a\neq 0$ 时。
\item  当 $a=0$ 时。这时有 $c\neq 0$.
\end{enumerate}

\end{itemize}

\end{frame}

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%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{8.4. 矩阵的初等变换 }

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\begin{itemize}

\item  习题4：设三阶行列式 $\det(A)=|a_{ij}|_{3\times 3} \neq 0$. 证明：存在一系列的行初等变换，使得
{\footnotesize 
\begin{eqnarray*}
A=\begin{bmatrix} 
a_{11} & a_{12} & a_{13} \\ 
a_{21} & a_{22} & a_{23} \\ 
a_{31} & a_{32} & a_{33} \\ 
\end{bmatrix}
\longrightarrow 
\begin{bmatrix} 
1&0&0 \\ 
0&1&0 \\ 
0&0&1 \\ 
\end{bmatrix}. 
\end{eqnarray*}
}

\item  解答思路：因为 $\det(A)\neq 0$, 所以矩阵 $A$ 的第一列不能全为零。

\end{itemize}

\end{frame}

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\begin{frame}{8.5. 矩阵的初等变换 }

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%每页详细内容

\begin{itemize}

\item  习题5：只用第三类（行或列）初等变换，实现下述矩阵的变换，
{\footnotesize 
\begin{eqnarray*}
A=\begin{bmatrix} 
1&2&3 \\ 
4&1&2 \\ 
3&4&1 \\ 
\end{bmatrix}
\longrightarrow 
\begin{bmatrix} 
1&0&0 \\ 
0&1&0 \\ 
0&0&d \\ 
\end{bmatrix}. 
\end{eqnarray*}
}

\item  解答思路：按标准步骤进行初等变换即得。注意到第三类（行或列）初等变换不改变行列式的值，所以一定有 $d=\det(A)$. 

\end{itemize}

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\begin{frame}[fragile=singleslide]{8.6. 矩阵的初等变换 }
%\begin{frame}{8.6. 矩阵的初等变换 }

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\begin{itemize}

\item  习题6：使用第一、二、三类行初等变换，和第一类列初等变换，实现下述矩阵的变换，
{\footnotesize 
\begin{eqnarray*}
A=\begin{bmatrix} 
1&1&2&5&7 \\ 
1&1&3&7&10 \\ 
1&1&4&9&13 \\ 
1&1&5&11&20 \\ 
\end{bmatrix}
\longrightarrow 
\begin{bmatrix} 
1&0&0&*&* \\ 
0&1&0&*&* \\ 
0&0&1&*&* \\ 
0&0&0&0&0 \\ 
\end{bmatrix}. 
\end{eqnarray*}
}

\item  解答思路：先化为行最简形。

%\begin{lstlisting}[language=R]
%A1 <- c(1,1,2,5,7); A2 <- c(1,1,3,7,10)
%A3 <- c(1,1,4,9,13); A4 <- c(1,1,5,11,20)
%A <- matrix(c(A1,A2,A3,A4), 4, 5, byrow = TRUE)
%rref(A)
%\end{lstlisting}

\end{itemize}

\end{frame}


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%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{8.7. 矩阵的秩 }

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%每页详细内容

\begin{itemize}

\item  习题7：使用初等变换求矩阵的秩：
{\footnotesize 
\begin{eqnarray*}
A=\begin{bmatrix} 
2&1&11&2 \\ 
1&0&4&-1 \\ 
11&4&56&5 \\ 
2&-1&5&-6 \\ 
\end{bmatrix}. 
\end{eqnarray*}
}

\item  解答思路：用行初等变换将矩阵 $A$ 化为行阶梯形。$R(A)=2$. 

\end{itemize}

\end{frame}

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\begin{frame}{8.8. 线性方程组有解的判别 }

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%每页详细内容

\begin{itemize}

\item  习题8：证明：若下述线性方程组有解，则其增广矩阵的行列式的值等于零，即：
{\footnotesize 
\begin{eqnarray*}
\left\{\begin{array}{rcl}
a_1x_1+a_2x_2+a_3x_3 &=& a_4 \\ 
b_1x_1+b_2x_2+b_3x_3 &=& b_4 \\ 
c_1x_1+c_2x_2+c_3x_3 &=& c_4 \\ 
d_1x_1+d_2x_2+d_3x_3 &=& d_4 \\ 
\end{array}\right.
\text{有解}
\hspace{0.3cm}
\Rightarrow 
\hspace{0.3cm}
\begin{vmatrix} 
a_1&a_2&a_3&a_4 \\  
b_1&b_2&b_3&b_4 \\  
c_1&c_2&c_3&c_4 \\  
d_1&d_2&d_3&d_4 \\  
\end{vmatrix} = 0. 
\end{eqnarray*}
}
 
\item  解答思路：不然的话，增广矩阵的秩就超过系数矩阵的秩了。

\end{itemize}

\end{frame}

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%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{8.9. 带参数的线性方程组 }

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%每页详细内容

\begin{itemize}

\item  习题9：求实数 $\lambda$ 的值，使得下述线性方程组有解：
{\footnotesize 
\begin{eqnarray*}
\left\{\begin{array}{rcl}
\lambda x_1 + x_2 + 2x_3 -3 x_4 &=& 2, \\ 
\lambda^2 x_1 -3 x_2 +2 x_3 + x_4 &=& -1, \\ 
\lambda^3 x_1 - x_2 +2 x_3 - x_4 &=& -1. \\ 
\end{array}\right.
\end{eqnarray*}
}

\item  解答思路：先将未知数 $x_1$ 与 $x_4$ 对换。然后将增广矩阵化为行阶梯形。必要时对 $\lambda$ 的不同取值进行讨论。

\end{itemize}

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%\begin{frame}[fragile=singleslide]{3.1.1. }
\begin{frame}{8.10. 带参数的线性方程组 }

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%每页详细内容

\begin{itemize}

\item  习题10：求实数 $\lambda$ 的值，分别使得下述线性方程组有唯一解、无解、无穷多解：
{\footnotesize 
\begin{eqnarray*}
\left\{\begin{array}{rcl}
\lambda x_1 + x_2 + x_3 &=& 1, \\ 
x_1 +\lambda  x_2 + x_3 &=& \lambda, \\ 
 x_1 + x_2 +\lambda x_3   &=& \lambda^2. \\ 
\end{array}\right.
\end{eqnarray*}
}

\item  解答思路：将增广矩阵化为行阶梯形。

\end{itemize}

\end{frame}

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\end{document}


